Question

Find the value of x and y using Cramer's rule:
3y = 3, 9x + 8y = 5x

Answer 1

Answer:

$3y=3,\:9x+8y=5x\quad :\quad y=1,\:x=-2$

Step-by-step explanation:

$\begin{bmatrix}3y=3\\ 9x+8y=5x\end{bmatrix}$

$\mathrm{Isolate\:solutions}$

$\mathrm{Isolate\:solution}:\quad 3y=3$

$3y+0\cdot \:x=3$

$\mathrm{Isolate\:solution}:\quad 9x+8y=5x$

$8y+4x=0$

$\begin{bmatrix}3y+0\cdot \:x=3\\ 8y+4x=0\end{bmatrix}$

$\mathrm{Matrix\:of\:Coefficients}$

$M=\begin{pmatrix}3&0\\ 8&4\end{pmatrix}$

$\mathrm{Answers\:column}$

$\begin{pmatrix}3\\ 0\end{pmatrix}$

$\mathrm{Replace\:the\:}y\mathrm{-column\:values\:with\:the\:answer-column\:values}$

$M_y=\begin{pmatrix}3&0\\ 0&4\end{pmatrix}$

$\mathrm{Replace\:the\:}x\mathrm{-column\:values\:with\:the\:answer-column\:values}$

$M_x=\begin{pmatrix}3&3\\ 8&0\end{pmatrix}$

$D:$

$M=\begin{pmatrix}3&0\\ 8&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$=3\cdot \:4-0\cdot \:8$

$=12$

$D_y$

$M_y=\begin{pmatrix}3&0\\ 0&4\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$=3\cdot \:4-0\cdot \:0$

$=12$

$D_x$

$M_x=\begin{pmatrix}3&3\\ 8&0\end{pmatrix}$

$\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc$

$=3\cdot \:0-3\cdot \:8$

$=-24$

$\mathrm{Solve\:by\:using\:Cramer\:Rule}$

$x=\dfrac{D_x}{D},\:y=\dfrac{D_y}{D},\:z=\dfrac{D_z}{D}$

$D\:\mathrm{denotes\:the\:determinat}$

$y=\dfrac{D_y}{D}=\dfrac{12}{12}$

$\mathrm{Simplify}$

$y=1$

$x=\dfrac{D_x}{D}=\dfrac{-24}{12}$

$\mathrm{Simplify}$

$x=-2$

$\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}$

$y=1,\:x=-2$